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O(ab) when o(a) and o(b) are relatively prime

  1. Apr 7, 2005 #1
    Hi,

    I am trying to prove that if o(a) and o(b) are relatively prime, and ab = ba, then o(ab) = o(a)o(b). I'd appreciate it if someone could give me a nudge in the right direction because I've spent almost 2 days on this now and I got nowhere. Which is rather annoying considering this is the first exercice in the chapter and the rest I did without a problem, so there must be something simple here that I'm missing. :mad:

    I already know that if (m, n) = 1 and m|k and n|k then mn|k. I think I can use this to prove what I need, if I can only show that o(a)|o(ab) and o(b)|o(ab). (Because I've already shown that o(ab)|o(a)o(b), so proving o(a)o(b)|o(ab) will be enough.)

    Thanks!
    Chen
     
    Last edited: Apr 7, 2005
  2. jcsd
  3. Apr 7, 2005 #2

    Hurkyl

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    What's o(x)? The order of x in the group of interest?

    I think a better approach would be to show o(ab) >= o(a) o(b)


    (P.S. I no longer think this approach is better!)
     
    Last edited: Apr 7, 2005
  4. Apr 7, 2005 #3
    Yes. Any idea how I could go about showing that? :smile:
     
  5. Apr 7, 2005 #4

    Hurkyl

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    o(a) | o(ab) sounds like Lagrange's theroem!

    As for o(ab) >= o(a) o(b), for some reason the division theorem springs to mind.
     
  6. Apr 7, 2005 #5
    I thought Lagrange's theorem deals with orders of groups, not orders of elements... :confused:
     
  7. Apr 7, 2005 #6

    Hurkyl

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    It does. That's called a hint. :smile:
     
  8. Apr 7, 2005 #7
    Sorry, I still don't get it. Which groups do you think should I define, to make use of Lagrange's theroem?

    I tried defining H = <ab>, Ka = <a> and Kb = <b>. But then H is a subgroup of KaKb and the only thing I can learn from that is that the order of H (which is o(ab)) divides the order of KaKb (which is o(a)o(b)) - but I already know this...
     
  9. Apr 7, 2005 #8
    And by the way, this chapter of questions comes before the chapter on Lagrange's theroem and even before the chapter on cyclic groups - so I think there's a way to solve this problem without making use of either of those.
     
  10. Apr 7, 2005 #9

    matt grime

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    Look at the group generated by ab, if a is in it, then you are done, or if yuou can show a^r is in it where r is some number coprime to o(a)....
     
  11. Apr 7, 2005 #10

    matt grime

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    If you don't like that method, then how about this (which is essentially the same):

    suppose that r is a positive integer and a^rb^r=e, then a^r=b^s, for some positive s (ko(b)-r for some multiple of o(b)). If we raise both sides to the power o(b), then a^(ro(b))=e from which it follows that o(a) divides r as o(a) and o(b) are coprime, and hence o(ab), by symmetry o(b) divides o(ab) and we are done.
     
  12. Apr 8, 2005 #11
    Thanks Matt, that did it. I took k=o(ab), then a^k=b^-k, raised to the power of o(b) etc.

    By the way, while we're on the subject, can you guys think of any groups of infinite order, in which every element is of finite order? The only one I found was the group of all roots of unity in the complex with regular multiplication. Are there any more? (well I guess there are a lot more...)
     
  13. Apr 8, 2005 #12

    Hurkyl

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    Take your favorite finite group, and take the group of all infinite sequences whose elements are in that group. (pointwise multiplication)
     
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